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What are all the exact solutions of [tex]-3 \tan^2 x + 1 = 0[/tex]? Give your answer in radians.

A. [tex]\frac{\pi}{6} + k \pi[/tex]
B. [tex]\frac{\pi}{6} + k \pi[/tex] and [tex]\frac{5\pi}{6} + k \pi[/tex]
C. [tex]\frac{\pi}{3} + k \pi[/tex] and [tex]\frac{5\pi}{3} + k \pi[/tex]
D. [tex]\frac{5\pi}{3} + k \pi[/tex]


Sagot :

To solve the equation [tex]\(-3 \tan^2(x) + 1 = 0\)[/tex], let’s go through the process step by step.

1. Isolate [tex]\(\tan^2(x)\)[/tex]:
[tex]\[ -3 \tan^2(x) + 1 = 0 \][/tex]
[tex]\[ -3 \tan^2(x) = -1 \][/tex]
[tex]\[ \tan^2(x) = \frac{1}{3} \][/tex]

2. Take the square root on both sides:
[tex]\[ \tan(x) = \pm \sqrt{\frac{1}{3}} \][/tex]
[tex]\[ \tan(x) = \pm \frac{1}{\sqrt{3}} \][/tex]
Simplifying, we get:
[tex]\[ \tan(x) = \pm \frac{\sqrt{3}}{3} \][/tex]

3. Identify the angles whose tangent values are [tex]\(\frac{\sqrt{3}}{3}\)[/tex] and [tex]\(-\frac{\sqrt{3}}{3}\)[/tex]:

[tex]\(\tan(x) = \frac{\sqrt{3}}{3}\)[/tex] occurs at [tex]\(x = \frac{\pi}{6} + k\pi\)[/tex], where [tex]\(k\)[/tex] is any integer.

[tex]\(\tan(x) = -\frac{\sqrt{3}}{3}\)[/tex] occurs at [tex]\(x = \frac{5\pi}{6} + k\pi\)[/tex], where [tex]\(k\)[/tex] is any integer.

4. Combine the two sets of solutions:
Therefore, the complete set of solutions to the equation [tex]\(-3 \tan^2(x) + 1 = 0\)[/tex] in radians is:
[tex]\[ x = \frac{\pi}{6} + k\pi \quad \text{and} \quad x = \frac{5\pi}{6} + k\pi \quad \text{for integer } k \][/tex]

Thus, the correct answer is:
[tex]\[ \boxed{\frac{\pi}{6} + k\pi \quad \text{and} \quad \frac{5\pi}{6} + k\pi} \][/tex]